Probability of error per bit = $10^{-7}$
Probability of bit arriving correctly = $1-10^{-7}$
Probabiilty of a 64 B frame arriving correctly = $(1-10^{-7})^{64*8} = 0.9999488 $
Probability of a frame being damaged = $1 – 0.9999488$ = $5.12 * 10^{-5}$
Since data-rate is 11-Mbps, no. of frames transmitted per second = $(11 * 10^{6}) / (64 * 8) = 21484.375$ frames/sec
No. of damaged frames per second = no. of frames/sec * prob. of a frame being damaged
$= 21484.375 * (5.12 * 10^{-5}) = 1.0999 = 1.1$ frames/sec (approx.)