Is my approach accurate to find original position of boat? A boat goes upstream for $3$ hr $30$ min and then goes downstream for $2$ hr $30$ min. If the speed of the current and the speed of the boat
in still water are $\frac{10}{3}$ kmph and $\frac{15}{2}$ kmph respectively, how far
from its original position is the boat now?

Speed of boat in still water$=\frac{15}{2}$
Speed of stream $=\frac{10}{3}$
Upstream speed $=\frac{15}{2}-\frac{10}{3}$
Downstream speed $=\frac{15}{2}+\frac{10}{3}$
→ Downstream distance - Upstream distance = far from original.
$=\left(2+\frac{1}{2}\right)\left(\frac{15}{2}+\frac{10}{3}\right) - \left(3+\frac{1}{2}\right)\left(\frac{15}{2}-\frac{10}{3}\right)$
$=\left(\frac{5}{2}\right)\left(\frac{65}{6}\right) - \left(\frac{7}{2}\right)\left(\frac{25}{6}\right)$
$=\frac{325-175}{12}$
→ $12.5$ km downstream.
Is my approach accurate to find original position of boat?

Your method looks great. As an alternative:  
With reference to ground, the water travels downstream for $2.5+3.5=6$ hours.
With reference to water, the boat went upstream for $1$ hour. So the boat is 
$$ \frac{10}{3} \times 6- \frac{15}{2}\times 1 = 12.5$$
$km$ downstream from the start. 
