If the $1^{st}$ drawn ball was $white$ then probability that $2^{nd}$ drawn ball is $white$ = $\left (\dfrac{m}{m+n}*\dfrac{m+k}{m+n+k} \right ) $ $\rightarrow (i)$
If the $1^{st}$ drawn ball was $black$ then probability that $2^{nd}$ drawn ball is $white$ = $\left (\dfrac{n}{m+n}*\dfrac{m }{m+n+k}\right ) $ $\rightarrow (ii)$
$(i)+(ii) = \left (\dfrac{m}{m+n}*\dfrac{m+k}{m+n+k} \right ) +\left (\dfrac{n}{m+n}*\dfrac{m}{m+n+k}\right )=\left (\dfrac{m}{m+n}\right ) *\left (\dfrac{m+n+k}{m+n+k}\right ) =\left (\dfrac{m}{m+n}\right )$
Option D is correct.