It should be option C.

Total lines = $\frac{2^{18}}{2^5}= 2^{13}$

Let x is set number.

Now whole address is of 32 bits out of which 16 + x + 5 (tagbits + set number + offset)

Hence x = 11 i.e $2^{11}$ sets will be there.

Hence k = $\frac{2^{13}}{2^{11}} = 4$

Total lines = $\frac{2^{18}}{2^5}= 2^{13}$

Let x is set number.

Now whole address is of 32 bits out of which 16 + x + 5 (tagbits + set number + offset)

Hence x = 11 i.e $2^{11}$ sets will be there.

Hence k = $\frac{2^{13}}{2^{11}} = 4$