$x\log_2 x$ is a continuously increasing function, and for a continuously increasing function $f(x)$,
$$\sum_{x=a}^{b} f(x) > \int_a^b f(x)dx$$
But in question, summation of L.H.S. above, $a=3$ and in R.H.S, $a=2$, so we don't know whether $S > T$. So we compute some initial values :
$\sum_{x=3}^{4} x\log_2 x \approx 12.754$, and $\int_2^4 x\log_2 x = 11$
Since $\sum_{x=3}^{4} x\log_2 x > \int_2^4 x\log_2 x$, and since we already know that
$\sum_{x=5}^{100} x\log_2 x > \int_5^{100} x\log_2 x$
So $\sum_{x=3}^{100} x\log_2 x > \int_2^{100} x\log_2 x$
So S > T, and option (A) is correct.