dear student, this ques is based directly on DESCARTEs KISSING CIRCLES THEOREM.
(you can google it and look up the wikipedia article for more info and refer the SPECIAL CASES section).
also, lets call 1/r1, 1/r2, …..., 1/rn as p(1), p(2),......., p(n)
by the theorem, we have
p(k)= 1+p(k-1) ± 2sqrt(1.p(k-1))= (1 ± sqrtp(k-1))^2 (coz for unit circle p= 1/r= 1/1= 1)
so sqrtp(k)= 1 ± sqrtp(k-1) now since clearly rk is less than r(k-1), we get that p(k-1) is less than p(k). also, each pk is greater than 1. taking these facts into consideration, it is easy to see that – ve sign is rejected. 
hence, sqrtp(k)= 1 + sqrtp(k-1) 
or p(k)= (1 + sqrtp(k-1))^2.
now, how do we find p(1)?
clearly, it would be p(1)= (1 + sqrtp(0))^2
but we havent defined p(0) till now. but we know from the theorem that it would be 1/r0 where r0 is radius of circle used for tangency to C1 apart from S1. hence r0 is radius of S2= 1
so, p(0)= 1/1= 1
so, p(1)= 4
similarly p(2)= (1+2)^2= 9 etc. u clearly observe the pattern p(k)= (k+1)^2 (you can rigorously prove this using induction too)
hence, Lt (p1+p2+.....+pn)/n^3
= Lt (4+9+.....(n+1)^2)/n^3
= Lt (1+4+9+.....(n+1)^2)/n^3 – Lt 1/n^3
= Lt (n+1)(n+1+1)(2(n+1)+1)/6n^3 – 0
= Lt (1+1/n)(1+2/n)(2+3/n)/6
= 1*1*2/6
= 1/3 = p/q
Hence p+q= 1+3
= 4
KINDLY APPROVE :))