Three grams each of magnesium and oxygen were allowed to react. Assuming that the reaction is complete to form $MgO$, the substance left unreacted is

No. of moles of magnesium $=$ weight $/MW=3/24=0.125moles$

No. of moles of oxygen $=$ weight $/MW=3/32=0.094moles$

The balanced equation for the formation of $MgO$ is

$2Mg+O_2⟶2MgO$

$2$ moles of magnesium react with $1$ mole of oxygen to form $2$ moles of $MgO$

$0.125$ moles of magnesium react with $0.125/2=0.0625$ moles of oxygen

From the above calculations it is concluded that oxygen is present in excess amount than magnesium. Magnesium is the limiting reagent

Moles of excess oxygen left = actual moles $-$ reacted moles $=0.094-0.0625=0.0315$moles.

From the formula Moles $=$ weight/MW, we can calculate the oxygen left unreacted $0.0315=W/32$

$W=0.0315\times 32=1.008 g$

Hence the correct option is $\left(D\right)$.