With in 1st 3 sec,kinetic energy will be #(1/2)m(v^2)# or #(1/2)m(u-g*t)^2# = #KE1# (u is the initial velocity)

Now,given within 6 sec it reaches the highest point,so initial velocity with which it was thrown was #g*t# or #(10*6)# or 60 m/s

So,KE1 = #(1/2)m(60-(10*3))^2# or #900(1/2)m#

At the highest point it's potential energy is #mgh# or #(mg*(u^2)/(2g))# or #(1/2)m(60)^2# or #3600(1/2)m# ; where, #h# is the maximum height reached by the ball. (i.e by the next 3 sec its kinetic energy got totally converted to potential energy)

Now,after 3 sec its potential energy was #mgh1# or #mg(((u^2)-(v^2))/(2g))# or #2700(1/2)m#( where, #h1# is the height reached in 3 sec)

So,with in these 3 sec increase in potential energy=decrease in kinetic energy=#KE2#=#(3600-2700)(1/2)m # or #900(1/2)m#

So, #KE1:KE2 = 1:1#