$(A)\;2 \\ (B)\;4 \\ (C)\;6 \\ (D)\;8 $

Original capacity $C_0= \large\frac{\in _09 A}{d}$

On introducing dielectric slab of thickness $\large\frac{d}{2}$ . the capacity becomes

$C= \large\frac{\in _0 A}{\Large\frac{d}{2k} + \bigg(d -\Large\frac {d}{2}\bigg ) }$

$\quad= \large\frac{ \in _0 A}{\Large\frac{d}{2} \bigg( \Large\frac{1}{K} +1\bigg) }$

As $c=\large\frac{4}{3}$

$\quad= \large\frac{ \in _0 A}{\Large\frac{d}{2} \bigg(1+ \Large\frac{1}{K} \bigg) }$

$\qquad= \large\frac {\in _0 A}{d} \times \large\frac{4}{3}$

=> $1+ \large\frac{1}{k}=\large\frac{3}{2}$

=> $\large\frac{1}{k} =\frac{1}{2}$

$K=2$

Hence A is the correct answer.

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