$\color{red}{\text{Without Replacement}}$

$\color{maroon}{\text{Conditional Probability} = \dfrac{\text{Number of Favourable Ways}}{\text{Total Number of Ways}}}$

Now, the urn having a total of $12$ balls in which $8$ balls are $\color{grey}{white}$.

So, remaining $(12-8)=4$ balls are $\color{black}{black}$.

Given that,

The sample space contains exactly $3$ $\color{grey}{\text{white balls}}$

& the favourable condition is - when the $1^{st}$ and $3^{rd}$ balls are of $\color{grey}{\text{white colour}}$.

Sample space contains exactly $3$ $\color{grey}{\text{white balls}}$

$S_1 = \{(\color{grey}{WWW}\color{black}{B},\color{grey}{WW}B\color{grey}{W},\color{grey}{W}B\color{grey}{WW},B\color{grey}{WWW})\}$

Sample space having $1^{st}$ and $3^{rd}$ balls as $\color{grey}{white}$.

$S_2=\{(\color{grey}{W}B\color{grey}{W}B),(\color{grey}{W}B\color{grey}{WW}),(\color{grey}{WWWW}),(\color{grey}{WWW}B)\}$

$∴S_{1}S_{2} = \{(\color{grey}{W}B\color{grey}{WW}),(\color{grey}{WWW}B)\}$

$∴ P(S_1) = \dfrac{8}{12}\times\dfrac{7}{11}\times\dfrac{6}{10}\times\dfrac{4}{9}+\dfrac{8}{12}\times\dfrac{7}{11}\times\dfrac{4}{10}\times\dfrac{6}{9}+\dfrac{8}{12}\times\dfrac{4}{11}\times\dfrac{7}{10}\times\dfrac{6}{9}+\dfrac{4}{12}\times\dfrac{8}{11}\times\dfrac{7}{10}\times\dfrac{6}{9}$

= $4\times\dfrac{4\times8\times7\times6}{12\times11\times10\times9}$

$∴ P(S_2) = \dfrac{8}{12}\times\dfrac{4}{11}\times\dfrac{7}{10}\times\dfrac{3}{9}+\dfrac{8}{12}\times\dfrac{4}{11}\times\dfrac{7}{10}\times\dfrac{6}{9}+\dfrac{8}{12}\times\dfrac{7}{11}\times\dfrac{6}{10}\times\dfrac{5}{9}+\dfrac{8}{12}\times\dfrac{7}{11}\times\dfrac{6}{10}\times\dfrac{4}{9}$

$∴ P(S_{1}S_{2}) = \dfrac{8}{12}\times\dfrac{4}{11}\times\dfrac{7}{10}\times\dfrac{6}{9} + \dfrac{8}{12}\times\dfrac{7}{11}\times\dfrac{6}{10}\times\dfrac{4}{9}$

= $2\times\dfrac{8\times4\times7\times6}{12\times11\times10\times9}$

$∴ \textbf{Required Probability} =$ $\dfrac{P(S_{1}S_{2})}{P(S_{1})} = \dfrac{2\times\dfrac{8\times4\times7\times6}{12\times11\times10\times9}}{4\times\dfrac{4\times8\times7\times6}{12\times11\times10\times9}}$

= $\dfrac{2}{4}$

= $\dfrac{1}{2}$