You won't be able to tell which effect is larger simply by looking at p values.
You haven't said how many people of each sex were in each group, but let's assume they are evenly split for now.
Let the outcome be $y=0$ for not cured, and $y=1$ for cured. The data are then
y Group Sex n
1 0 Group A Female 460
2 0 Group A Male 430
3 0 Group B Female 450
4 0 Group B Male 445
5 1 Group A Female 40
6 1 Group A Male 70
7 1 Group B Female 50
8 1 Group B Male 55
Let's model the risk of being cured using
$$ \log \left( \dfrac{p}{1-p} \right) = \beta_0 + \beta_1x_1 + \beta_2 x_2 + \beta_3x_1x_2 $$
Here:
- $\beta_0$ is the reference category (females from group A).
- $\beta_1$ is the effect of being in group B
- $\beta_2$ is the effect of being male
- $\beta_3$ is the interaction of group B and being male.
Hence, the risk of being cured on the logit scale from group A is
$$\log \left( \dfrac{p}{1-p} \right) = \beta_0 + \beta_2$$
and the risk of being cured on the logit scale from group B is
$$\log \left( \dfrac{p}{1-p} \right) = \beta_0 + \beta_1 + \beta_2 + \beta_3$$
The difference in risks is the quantity $\beta_1 + \beta_3$. Our null is that $\beta_1 + \beta_3 = 0$ versus the alternative that $\beta_1 + \beta_3 \neq 0$. Using R to perform the computations...
# Fit a logistic regression to the data
mod<-glm(y ~ Group*Sex, data = model_data, family=binomial(), weights = n)
# Extract the covariance matrix for the coefficients
Sigma = vcov(model)
# Compute the standard error of the estimate using the covariance matrix
x = c(0, 1, 0, 1)
b = sum(coef(mod) %*% x)
se_b = x %*% Sigma %*% x
# Test statistic
z = b/sqrt(se_b)
>>> -1.43
The test statistic is $z = -1.43$ yielding a p value of about 0.15. We would fail to reject the null hypothesis that males have different risks of being cured across populations.
EDIT:
There may be a simpler way of testing this. The stated hypothesis is really about homogeneity of odds ratios between strata, here stratified by population. We can do this using Cochran's test of homogeneity. I'll work with log odds ratios because they are slightly simpler algebraically.
Let $\hat{\theta}_k$ be the log odds ratio in the $k^{th}$ strata. Let $\hat{\theta}$ be the estimate of the marginal odds ratio (pooling strata). Asymptotically,
$$\widehat{\theta}_{k}-\theta \stackrel{d}{\approx} N\left(0, \sigma_{\widehat{\theta}_{k}}^{2}\right)$$
Let $\tau_k = \sigma^{-2}_{\hat{\theta}_k}$ be the estimated precision of the log odds ratio. A test statistic for homogeneity of odds ratios is
$$X_{H, C}^{2}=\sum_{k} \widehat{\tau}_{k}\left(\widehat{\theta}_{k}-\hat{\theta}\right)^{2}$$
Where $X^2_{H, C} \sim \chi^2_{K-1}$. When I compute this test statistic, I get $X^2_{H, C} = 3.136$ which yields a p value of 0.076, again failing to reject the null. For more, see section 4.6.2 of this book.