Miss Penalty = Main memory access time + Cache load time
= 10 + 17 = 27 cycles. -- assuming this is same for both data and instruction cache
Instruction Execution Time with cache (both instruction and cache) ignoring data memory access = Instruction Cache access time + Miss Rate * Miss penalty
$= 1ns + 0.05 \times 27 = 2.35 ns$.
Average data memory access time = Data cache access time + Miss rate * Miss penalty
$ = 1ns + 0.1 \times 27 = 3.7 ns$.
Average instruction execution time $ = 2.35 + 0.3 \times 3.7 = 3.46 ns$.
Average instruction execution time without cache $ = 10 + 0.3 \times 10 = 13 ns$.
So, speed-up $ = \frac{13}{3.46} = 3.75.$