The one-way ANOVA model is

$Y_{ij}=\mu_i+\epsilon_{ij}, \qquad \epsilon_{ij} *i.i.d.* N(0,\sigma^{2}),$

and the hypothesis tested is $H_0:\mu_1=\mu_2=\cdots=\mu_k$ .

The three "standard assumptions" are independence, normality, and — critically here — homoscedasticity: a common variance $\sigma^{2}$ across all groups. That common $\sigma^{2}$ is what the error mean square estimates, and it is the single yardstick in the denominator of

$F=\frac{\mathrm{MS}_{\text{Between}}}{\mathrm{MS}_{\text{Within}}} .$

If the group variances differed, $\mathrm{MS}_{\text{Within}}$ would be pooling unlike quantities and the $F$ reference distribution would no longer apply.

So ANOVA tests equality of means under the assumption that the variances are all equal — option (B), which states the hypothesis together with the condition under which the test is valid. Option (A) is incomplete; (C) describes Bartlett's or Levene's test; and equality of variances in (D) is an assumption, not part of the hypothesis being tested.