From Euler Formula We know for connected graph having $v$ vertices ,$e$ edges,$f$ faces ,
$v-e+f=2$.
In problem it is given ,
$v=10$
Given graph is bipartite , which implies The is no odd length cycle in the graph => There is no triangle in the graph => every face must bounded by at least 4 edges.
And we know that, degree summation of each face= $2*e$.
So we can write ,
$4f <=2e$
or, $2f<=e$
or, $2(2+e-v)<=e$
or ,$4+2e-2v<=e$
or,$e<=2v-4$
Here it is given $v=10$
So, maximum number of edges possible , $2*10-4=16$
So ,Atmost 16 edges possible .