Answer
A has resent $\mathrm{M}(3),$ presumably because $\mathrm{B}^{\prime} \mathrm{s\ ACK}(3)$ message never
reached A. When $\mathrm{B}$ receives the second copy of $\mathrm{M}(3),$ it should again
send the $\mathrm{ACK}(3)$ to $\mathrm{A}$ but discard the message because it is a duplicate.
Work Step by Step
A has resent $\mathrm{M}(3),$ presumably because $\mathrm{B}^{\prime} \mathrm{s\ ACK}(3)$ message never
reached A. When $\mathrm{B}$ receives the second copy of $\mathrm{M}(3),$ it should again
send the $\mathrm{ACK}(3)$ to $\mathrm{A}$ but discard the message because it is a duplicate.
