The present value of an income stream with $n$ equally spaced payments, with an interest of $r$ at each payment, each payment of $C$, is obtained by computing the present value of each payment and adding it up. Here, you have $r=0.06/4 = 0.015$, as you note, so you get $$1000 + \frac{1000}{1+r} + \frac{1000}{(1+r)^2} + \cdots + \frac{1000}{(1+r)^{100}},$$ since there are 100 payments (one payment per quarter, 25 years, beginning now).
This is just $$1000\left(1+\frac{1}{1+r} + \left(\frac{1}{1+r}\right)^2 + \cdots + \left(\frac{1}{1+r}\right)^{100}\right)$$ which is $$1000\left( \frac{1 - \frac{1}{(1+r)^{101}}}{1-\frac{1}{1+r}}\right).$$
If you only get 1000 payments (no payment at the end of the last quarter 25 years from now) then the sum only goes up to exponent 99; adjust accordingly.