Question 34.

In any right triangle, the circumradius is half of the hypotenuse.

Here,  

$L = \frac{1}{2} \times \text{(length of the hypotenuse)}$

$L = \frac{1}{2}\left(\sqrt{15^{2} + 9^{2}}\right)$

= $\frac{1}{2}\sqrt{306}$

= $\frac{1}{2} \times 17.49$

= $8.74$

Hence, the integer close to $L$ is $9$.