Let event $\mathrm{A}=$ Picked card is a queen. and event $\mathit{B}=$ card of spade is picked.
$\mathrm{P}\left(\mathrm{A}\right)=\frac{4}{52},\mathrm{P}\left(\mathrm{B}\right)=\frac{13}{52}$
and, $\mathit{P}\left(\mathit{A}\cap \mathit{B}\right)=\frac{1}{52}[\because \mathit{A}\cap \mathit{B}=$ Queen of spade $]$
Now, required probability,
$\mathrm{P}\left(\mathrm{B}∕\mathrm{A}\right)=\frac{\mathit{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{A}\right)}=\frac{1∕52}{4∕52}=\frac{1}{4}$
$\therefore$ Option (c) is correct.