Let the speed of the river be $S_{r}$ and the speed of the boat be $S_{b}.$
Let $d$ be the distance and $t$ be the initial time taken.
We know that, $\boxed{\text{Time} = \frac{\text{Distance}}{\text{Speed}}}$
Now, $t = \underbrace{\frac{d}{S_{b}+S_{r}}}_{\text{Downstream time}} + \underbrace{\frac{d}{S_{b}-S_{r}}}_{\text{Upstream time}} \quad \longrightarrow (1)$
If he doubles the speed of his motorboat, then his total travel time gets reduced by $75\%$.
So, $t\times \frac{75}{100} = \frac{d}{2S_{b}+S_{r}} + \frac{d}{2S_{b}-S_{r}}$
$\Rightarrow \frac{t}{4} = \frac{d}{2S_{b}+S_{r}} + \frac{d}{2S_{b}-S_{r}} $
$\Rightarrow t = \frac{4d}{2S_{b}+S_{r}} + \frac{4d}{2S_{b}-S_{r}} \quad \longrightarrow (2)$
Equate the equation $(1)$ and equation $(2),$ we get.
$\frac{d}{S_{b}+S_{r}} + \frac{d}{S_{b}-S_{r}} = \frac{4d}{2S_{b}+S_{r}} + \frac{4d}{2S_{b}-S_{r}}$
$\Rightarrow d \left[\frac{S_{b}-S_{r}+S_{b}+S_{r}}{(S_{b}+S_{r})(S_{b}-S_{r})}\right] = 4d\left[\frac{2S_{b}-S_{r}+2S_{b}+S_{r}}{(2S_{b}+S_{r})(2S_{b}-S_{r})}\right] $
$\Rightarrow \frac{2S_{b}}{S_{b}^{2} – S_{r}^{2}} = \frac{16S_{b}}{4S_{b}^{2} – S_{r}^{2}} $
$\Rightarrow 4S_{b}^{2} - S_{r}^{2} = 8S_{b}^{2} - 8S_{r}^{2}$
$\Rightarrow 7S_{r}^{2} = 4S_{b}^{2} $
$\Rightarrow 4S_{b}^{2} = 7S_{r}^{2} $
$\Rightarrow \frac{S_{b}^{2}}{{S_{r}^{2}}} = \frac{7}{4}$
$\Rightarrow \left(\frac{S_{b}}{S_{r}}\right)^{2} = \frac{7}{4}$
$\Rightarrow\frac{S_{b}}{S_{r}} = \sqrt{\frac{7}{4}}$
$\Rightarrow\frac{S_{b}}{S_{r}} = \frac{\sqrt{7}}{2}$
$\therefore$ The ratio of the original speed of the motorboat to the speed of the river is $\sqrt{7}: 2.$
Correct Answer $:\text{B}$