Speed in downstream $ = S_{b} + S_{w},$ Speed in upstream $ = S_{b} - S_{w},$ where $S_{b} = $ speed of the boat in still water, $S_{w} = $ speed of the stream.
We know that $\text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \implies \text{Time} = \dfrac{\text{Distance}}{\text{Speed}}$
Given that, $S_{b} = 10\;\text{mph}$
Now, Speed in downstream $ = (10 + S_{w})\;\text{mph},$ Speed in downstream $ = (10 - S_{w})\;\text{mph}$
$\implies \dfrac{36}{10 – S_{w}} – \dfrac{36}{10+S_{w}} = \dfrac{90}{60}$
$\implies \dfrac{36(10+S_{w}) – 36(10-S_{w})}{(10-S_{w})(10+S_{w})} = \dfrac{9}{6} = \dfrac{3}{2}$
$\implies \dfrac{360 + 36S_{w} – 360 + 36S_{w}}{100 – S_{w}^{2}} = \dfrac{3}{2}$
$\implies 144S_{w} = 300 – 3S_{w}^{2}$
$\implies 48S_{w} = 100 – S_{w}^{2}$
$\implies S_{w}^{2} + 48S_{w} – 100 = 0$
$\implies S_{w}^{2} + 50S_{w} – 2S_{w} – 100 = 0$
$\implies S_{w}(S_{w} + 50) – 2(S_{w} + 50) = 0$
$\implies (S_{w} – 2)(S_{w} + 50) = 0$
$\therefore S_{w} = 2\;\text{mph}, \;S_{w} \neq -50\;\text{mph}$
So, the correct answer is $(A).$