Let me just introduce some notation first to make this easier. Let's denote the total number of data points in the full sample as $N$. The variance of the full sample is
$$\sigma_N^2 = \frac{1}{N} \sum_{n=0}^{N} (x_i - \mu_N)^2$$
for the total sample mean $\mu_N$, where you might change the normalisation factor to $N-1$ for an unbiased estimator. Similarly, we will denote the subsample variances and means as $\sigma_A^2$, $\sigma_B^2$ and $\mu_A$, $\mu_B$ and we'll say that the number of draws in subsample $A$ is $n_A$, leaving subsample $B$ with $n_B = N - n_A$.
If we expand the above expression, we get
$$\sigma_N^2 = \frac{1}{N} \sum_{n=0}^N(x_i^2 - 2\mu_Nx_i + \mu_N^2)$$
and bringing the sum through gives us
$$\sigma_N^2 = \frac{1}{N} \sum_{n=0}^{N} x_i^2 - 2\mu_N \left( \frac{1}{N} \sum_{n=0}^{N} x_i \right) +\frac{1}{N}\sum_{n=0}^{N}\mu_N^2 $$
If we add $\mu_N^2$ to both sides we are left with
$$\sigma_N^2 + \mu_N^2 = \frac{1}{N} \sum_{n=0}^N x_i^2$$
which we can do similarly with the subsamples. Splitting the sum in two over the subsamples, we now get
$$\sigma_N^2 + \mu_N^2 = \frac{1}{N} \left[\sum_{n=0}^{n_A} x_i^2 + \sum_{n=n_m + 1}^{N}x_i^2 \right]$$
and we can subsititute the sums for the identity derived above in each subsample
$$\sigma_N^2 + \mu_N^2 = \frac{1}{N} \left[ n_A \left( \sigma_A^2 + \mu_A^2 \right) + n_B \left( \sigma_B^2 + \mu_B^2 \right) \right] $$
So finally, after rearranging, we can get the variance of the second subsample as a function of the means and other variances:
$$\sigma_B^2 = \frac{N\left(\sigma_N^2 + \mu_N^2\right) - n_A \left(\sigma_A^2 + \mu_A^2 \right)}{n_B} - \mu_B^2$$
This will be slightly different if you are using unbiased estimators for the variance. You can see extra info relating to this problem in answers to the following other questions:
How to calculate the variance of a partition of variables
How to calculate pooled variance of two groups given known group variances, means, and sample sizes?