Which among the countries P, X, and C has/have the least total trade?

In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.
In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.
For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.
For X,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$
$10E\ -\ 10I\ =\ E\ +\ I$
$11I\ =\ 9E$
So, let us assume export to be 11b and import to be 9b.
For C,
$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$
$5I\ -\ 5E\ =\ E\ +\ I$
$2I\ =\ 3E$
So, let us assume export to be 2c and import to be 3c.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.
40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P
22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X
Equating both, we get,
4.4b = 2.2a
a = 2b.
We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.
The total imports for C = 3c = 1200
c = 400
This makes the exports for C = 2c = 800
In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.
90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P
4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW
The exports from C to X = 800 - 720 - 32 = 48
In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.
If we assume that Imports of ROW are m and exports of ROW are n.
12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X
40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P
Placing all the values in the table, we get,
Equating the imports of P and X to the Totals, we get,
For P,
4.4b + 720 + 0.4n = 20b
15.6b = 720 + 0.4n --(1)
For X,
600 + 48 + 0.12n = 9b
9b = 648 + 0.12n --(2)
Solving (1) and (2), we get
$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$
$10108.8+1.872n=6480+3.6n$
$3628.8=1.728n$
$n=2100$
From (1),
15.6b = 720 + 0.4(2100)
15.6b = 1560
b = 100
Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200
Exports of X to ROW = 11b - 4.4b = 6.6b = 660
Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.
The value of exports and imports of ROW to ROW has to be equal.
We calculated the value of n to be 2100.
The exports of ROW to ROW can be calculated as,
Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.
The value of m can be calculated as,
m = 200 + 660 + 32 + 1008 = 1900.
Filling up the table with all the values, we get,
Total trade of P = 2000 + 2000 = 4000
Total trade of X = 900 + 1100 = 2000
Total trade of C = 1200 + 800 = 2000
So, both countries X and C have the least total trade.
Hence, the correct answer is option C.