The last instruction of the given code modifies the register$/$memory

$s$3$

to the value $0$x

$0$

$.$

Note: The value should be hexadecimal $($do not write $0$x$)$ without any leading zeros.

Code:

lbu $t$2,1($$s$0)$

xor $s$3,$ $t$8,$ $t$2$