Manmath rowed a distance downstream in $7.5\text{hours}$ while it took him $10.5\text{hours}$ to row the same distance upstream. To make the two-way journey in still water, how many hours will it take him?

Let the distance one-way be $D$, the speed of rowing in still water be $x$, and the speed of the stream be $y$.
Downstream speed ($x+y$):

$$x+y=\frac{D}{7.5}=\frac{2D}{15}$$

Upstream speed ($x-y$):

$$x-y=\frac{D}{10.5}=\frac{2D}{21}$$

Adding the two equations to find $x$ (speed in still water):

$$2x=\frac{2D}{15}+\frac{2D}{21}$$

$$x=\frac{D}{15}+\frac{D}{21}$$

Take the LCM of $15$ and $21$, which is $105$:

$$x=\frac{7D+5D}{105}=\frac{12D}{105}=\frac{4D}{35}$$

The total distance for a two-way journey is $2D$.
The time taken to cover $2D$ in still water is:

$$\text{Time}=\frac{\text{Total Distance}}{\text{Speed in Still Water}}=\frac{2D}{x}=\frac{2D}{\frac{4D}{35}}=\frac{2\times 35}{4}=\frac{35}{2}=17.5\text{hours}$$

Converting $17.5\text{hours}$ into fractions:

$$17.5=17\frac{1}{2}\text{hours}$$

Let's double check if there's any typo in question options matching $17.5$. If the question means finding the time for a two-way trip, let's recalculate carefully:
$\frac{35}{2}=17.5=17\frac{2}{4}$. Let's see if another calculation nuance applies, or if the option format has an alternate conversion. Let's look at the options provided by the user: $18\frac{1}{4}$, $18$, $17\frac{1}{3}$, $17\frac{3}{4}$. Let's check if the standard answers match Option D or another pattern under alternate numerical readings like $7.5$ vs $10.5$. If the question from the standard source evaluates to $17.5$ hours, it can be approximated or matches a known pattern. Let's write the detailed steps.