Answer
$Ni(CO)_4$
Work Step by Step
Number of moles of $CO_2$:
$0.100\ g\div 44.001\ g/mol=0.00227\ mol$
$0.00227\ mol$ of CO
Number of moles of $NiO$:
$0.0426\ g\div 74.693\ g/mol=0.00057\ mol$
$0.00057\ mol$ of Ni
Ratio: $4.00$
Empirical formula:
$Ni(CO)_4$