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The magnetic flux is given by
$\Phi =BA\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=B\left(\pi {r}^{2}\right)\mathrm{cos}\theta $
Substituting the values and solving gives
$\Phi =\left(0.078T\right)\left(\pi {\left(0.10m\right)}^{2}\right)\mathrm{cos}25\xb0\phantom{\rule{0ex}{0ex}}=2.22\times {10}^{-3}Wb$