1. Given

   • Mass of block $= m$
   • Platform starts from rest
   • Upward acceleration of platform $= \dfrac{g}{2}$
   • We need work done by normal reaction on the block in time $t$.

2. Forces on the block The block remains in contact with the accelerating platform, so it also accelerates upward with $a = \frac{g}{2}.$

   Forces on the block:

   • Upward normal reaction $N$
   • Downward weight $mg$

   Applying Newton's second law upward: $N - mg = m\left(\frac{g}{2}\right)$ $N = mg + \frac{mg}{2} = \frac{3mg}{2}.$

3. Displacement of the block in time $t$ Initial velocity is zero, acceleration is $\dfrac{g}{2}$ upward.

   So displacement in time $t$ is $s = 0\cdot t + \frac{1}{2}\left(\frac{g}{2}\right)t^2 = \frac{gt^2}{4}.$

4. Work done by normal reaction Normal reaction and displacement are both upward, so $W_N = Ns.$

   Substituting: $W_N = \left(\frac{3mg}{2}\right)\left(\frac{gt^2}{4}\right)$ $W_N = \frac{3mg^2t^2}{8}.$

5. Match with options This corresponds to: $\boxed{\frac{3mg^2t^2}{8}}$ which is Option B.

6. Check sign in the question The statement says "Work done by normal reaction on block in time is -" which appears to mean "is equal to" rather than asking for a negative quantity. Since force and displacement are in the same direction, the work by normal reaction is positive, not negative.

Therefore, the correct answer is B.