If $v_{w}$ be the velocity of water and $v_{b}$ be the velocity of motorboat in still water.
The distance covered by motorboat in moving downstream in $6\, h$ is
$x=\left(v_{b}+v_{w}\right) \times 6$...(i)
Same distance covered by motorboat in moving upstream in $10\, h$ is
$x=\left(v_{b}-v_{w}\right) \times 10$...(ii)
From Eqs. (i) and (ii), we have
$\left(v_{b}+v_{w}\right) \times 6 =\left(v_{b}-v_{w}\right) \times 10$
$v_{w} =\frac{v_{b}}{4}$
$\therefore x =\left(v_{b}+v_{w}\right) \times 6=7.5\, v_{b}$
Time taken by the motorboat to cover the same distance in still water is
$t=\frac{x}{v_{b}}=\frac{7.5 v_{b}}{v_{b}}=7.5\, h$