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22 < 2^5, so 22^99 < 2^495 < 2^999

99 < 2^7, so 99^22 < 2^154 < 2^999

999 < 2^10, so 999^2 < 2^20 < 2^999

Let 2 $^{999}$ =x

Log 2 $^{999}$ =logx

999log2=logx

300.699=logx

10 $^{300.699}$ =x

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Let 22 $^{99}$ =y

log22 $^{99}$ =logy

99log22=logy

132.89=logy

10 $^{132.89}$ =y

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Let 99 $^{22}$ =z

Log99 $^{22}$ =logz

22log99=logz

43.77=logz

10 $^{43.77}$ =logz

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Let 999 $^2$ =w

Log 999 $^2$ =logw

2log999=logw

5.998=logw

10 $^{5.998}$ =w

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Hence, 2 $^{999}$ is greatest

A. 5357543035931336604742125245300009052807024058527668037218751941851755255624680612465991894078479290637973364587765734125935726428461570217992288787349287401967283887412115492710537302531185570938977091076523237491790970633699383779582771973038531457285598238843271083830214915826312193418602834034688

B.
7940455229131020726743381854098636404416494857253597834213839631812461973959829138671244625892339275269942476459642840306880094404608

C.
80163058953904597655850360719950389962307801

D.
998001

$2^{999}>2^{10 \times 99}=(2^{10})^{99}>22^{99}$ , $(2^{10})^{99}>99^{22}$ , $(2^{10})^{99}>999^2$ . biggest number is $2^{999}$ !