The divisors of the given number will have the form $2^a*3^b*5^c*7^d$ with $0\le a\le6,\ 0\le b\le5,\ 0\le c\le3,\ 0\le d\le2$
Because the divisors should be in the form 3r+1, it cannot be divisible by 3, so (b=0).
Reduce modulo $2 \bmod 3 = 2,\; 5 \bmod 3 = 2,\; 7 \bmod 3 = 1$
Hence, $2^a5^c7^d\equiv 2^{a+c}\cdot1^d \equiv 2^{a+c}\pmod3$
2^k will be in the form 3r+1 only when K is even. So, we need a+c to be even
$a\in{0,\dots,6}$ has 4 even, 3 odd values
$c\in{0,\dots,3}$ has 2 even, 2 odd
Number of (a,c) with (a+c) even is $4\cdot2 + 3\cdot2 = 8+6=14$
For each such pair, there are 3 choices for d = 0,1,2. Thus, total divisors in the form (3r+1) equals $14\times3=42$.
