Answer
$$f''(2)=200$$
Work Step by Step
$f(2)=10$ and $f'(x)=x^2f(x)$ for all $x$
Find $f''(x)$ $$f''(x)=(x^2)'f(x)+x^2f'(x)$$ $$f''(x)=2xf(x)+x^2f'(x)$$ $$f''(x)=2xf(x)+x^2(x^2f(x))$$ $$f''(x)=2xf(x)+x^4f(x)$$
So, $$f''(2)=2\times2f(2)+2^4f(2)$$ $$f''(2)=4\times10+16\times10$$ $$f''(2)=200$$