I assume that there is duplication in the typing of this question, and the river speed is 6kph (6km/h).
When the boat travels upstream its effective speed is v-6 and its downstream speed is v+6 where v is the boat’s speed in still water.
53=(v-6)t₁=(v+6)t₂ and t₁+t₂=15. So t₂=15-t₁.
(v-6)t₁=(v+6)(15-t₁)=53.
vt₁-6t₁=15v+90-vt₁-6t₁ so t₁=(90+15v)/2v.
(v-6)(90+15v)=106v, 90v-540+15v²-90v=106v.
15v²-106v-540=0; v=(106±√106²+32400)/30=10.496kph approx.