Answer
$One\ solution\ is\ as\ follows:$
$15E0 \rightarrow$ (LOAD register 5 with the contents of memory cell E0.)
$A502 \rightarrow$ (ROTATE the contents of register 5 to the left by 2 bits.)
$260F \rightarrow$ (LOAD register 6 with the value 0F.)
$8056 \rightarrow$ (AND registers 5 and 6, leaving the result in register 0.)
$30E1 \rightarrow$ (STORE the contents of register 0 in memory cell E1.)
$C000 \rightarrow$ (HALT.)
Work Step by Step
$One\ solution\ is\ as\ follows:$
$15E0 \rightarrow$ (LOAD register 5 with the contents of memory cell E0.)
$A502 \rightarrow$ (ROTATE the contents of register 5 to the left by 2 bits.)
$260F \rightarrow$ (LOAD register 6 with the value 0F.)
$8056 \rightarrow$ (AND registers 5 and 6, leaving the result in register 0.)
$30E1 \rightarrow$ (STORE the contents of register 0 in memory cell E1.)
$C000 \rightarrow$ (HALT.)