Error in trapezoidal rule is given by : $$E_n = -\frac{(b-a)^3}{12N^2} f''(c)$$ for some value $c$ between $a$ and $b$. Now, we want accuracy of at least $\frac{1}{3}*10^{-6}$ meaning maximum error will be $\frac{1}{3}*10^{-6}$ i.e. $$\left\vert E_n\right\vert < \frac{1}{3}*10^{-6}$$ We are given $a$ as 1 and $b$ as 2, and we need to find $N$. By rearranging terms, we get $$N^2 > \frac{(b-a)^3}{12*\frac{1}{3}*10^{-6}}\left\vert f''(c) \right\vert = \frac{(2-1)^3}{4*10^{-6}}\left\vert f''(c) \right\vert = \frac{10^6}{4}\left\vert f''(c) \right\vert$$ So $$N > \frac{10^3}{2}\sqrt{\left\vert f''(c) \right\vert}$$ Now $f''(x) = xe^x + 2e^x$, and between 1 and 2, it can take maximum value at 2 (we are looking at max value because we want R.H.S. of above inequality to be as large as poosible so that we can find least bound on N). So $f''(2) = 4e^2$. So $$N > \frac{10^3}{2}\sqrt{4e^2} = \frac{10^3}{2}*2e = 1000e$$ Hence, option (A) is correct.
