We have 32 bit instruction
We need 7 bits for specifying an particular register address ($\left \lceil log_{2}100 \right \rceil$)
We need 20 bits to address a particular word in memory $\left \lceil log_{2}1 M \right \rceil$
out of 32 if we subtract k and L address would get 0 address instruction
$2^{32}-k*2^{7}*2^{20}-L*2^{7}$
$((2^{5}-k)2^{20}-L)2^{7}$