Yes - if you're measuring a proportion, and p=0.4 is the population mean, then your population standard deviation is sqrt(p(1-p)) = 0.49. If your true population mean is unknown, you'd use p=0.5 (which gives you the maximum sd, 0.5).
Then you find the critical z for a 95% confidence interval (1.96), then you know 0.01 = 1.96 x SEM.
This gives you a required standard error (SEM) of 0.01/1.96 = 0.0051, and since SEM = sd / sqrt(n) we can rearrange to give:
n = (sd / SEM)^2
= (0.5 / 0.0051)^2 [ or use 0.49 / 0.0051 if you know p=0.4 in your population)