To solve this problem, we need to determine how many cupcakes can be purchased for the same amount that Lena spent, given the prices of each pastry type.

Let's denote the prices:
- The price of one cake as $c$.
- The price of one cupcake as $p$.
- The price of one bagel as $b$.

According to the problem, Petya, Apya, and Kolya all spent the same amount of money, leading to the following equations:
- Petya: $c + 2p + 3b$
- Apya: $3c + b$
- Kolya: $6p$

These equations imply:
$$c + 2p + 3b = 3c + b = 6p$$

We want to find the number of cupcakes $n$ Lena could buy with the same amount of money she spent on 2 cakes and 2 bagels:
- Lena: $2c + 2b$

First, set the expressions for Petya and Apya equal:
$$c + 2p + 3b = 3c + b$$
Simplifying, we have:
$$2p + 2b = 2c$$
$$p + b = c$$

Next, equate the expressions for Petya and Kolya:
$$c + 2p + 3b = 6p$$
Simplifying, we find:
$$c + 3b = 4p$$

We now express $2c + 2b$ in terms of $p$. Using $p + b = c$, substitute $c = p + b$ into expression for Lena:
$$2(p + b) + 2b = 2p + 2b + 2b = 2p + 4b$$

Now, relate $2p + 4b$ to $6p$. From $c + 3b = 4p$, substitute $c = p + b$ to get:
$$p + b + 3b = 4p \quad \Rightarrow \quad p + 4b = 3p \quad \Rightarrow \quad 2b = 2p - p \quad \Rightarrow \quad p = 2b$$

Substitute $p = 2b$ in Lena's equation:
$$2p + 4b = 4b + 4b = 8b$$

Finally, equating $8b$ to the equated amounts $6p$, notice:
$$8b = 6p \quad \Rightarrow \quad 8b = 6(2b) \quad \Rightarrow \quad p = \frac{4}{3}b$$

Thus, Lena could purchase $\frac{13}{4}$ cupcakes for the amount she spent. Therefore, the number of cupcakes Lena could buy is:
$$\boxed{\frac{13}{4}}$$