Let
x = rate of boatman in still water
y = rate of stream
A man can row 77 km and back in 14 hours
$t_{\text{going downstream}} + t_{\text{going upstream}} = 14$
$\dfrac{77}{x + y} + \dfrac{77}{x - y} = 14$
$\dfrac{11}{x + y} + \dfrac{11}{x - y} = 2$ ← Equation (1)
He can row 6.5 km with the stream at the same time as 4.8 km against the stream
$t_{\text{6.5 km with the stream}} = t_{\text{4.8 km against the stream}}$
$\dfrac{6.5}{x + y} = \dfrac{4.8}{x - y}$
$6.5(x - y) = 4.8(x + y)$
$6.5x - 6.5y = 4.8x + 4.8y$
$1.7x = 11.3y$
$x = \frac{113}{17}y$
Substitute x = 113y/17 to Equation (1)
$\dfrac{11}{\frac{113}{17}y + y} + \dfrac{11}{\frac{113}{17}y - y} = 2$
$\dfrac{11}{\frac{130}{17}y} + \dfrac{11}{\frac{96}{17}y} = 2$
$\dfrac{187}{130y} + \dfrac{187}{96y} = 2$
$\dfrac{187}{130} + \dfrac{187}{96} = 2y$
$y = 1.693 ~ \text{kph}$ answer