Well, here is an approach that justifies the answer of 5.5. I'm not really up on the terminology. Given that, I think the exact match approach is not possible because there is not a bond that pays off in two periods so interest rate changes may preclude an exact match.
For immunization, we want to form a portfolio of bonds that delivers $\$1000$ in period two, and whose value is not very sensitive to changes in interest rates. Let $x$ be the number of one-period bonds we buy, and $y$ be the number of three-period bonds we buy (each with face value 100).
The value of this portfolio in period two is
$$V=x(1+R_{12})100 +\frac{y}{(1+R_{12})}100.$$
Here $R_{12}$ is the interest rate between period 1 period 2 (today is 0, just to be clear). That is, you buy a one period bond today, it gives you 100 tomorrow, and you roll it over from 1 to 2, and it gives you $100(1+R_{12})$ though as of today we don't know what that will be. The value of the three-period bond in period 2 should be described as $\frac{y}{(1+R_{23})}100$, where $R_{23}$ is the interest rate between 2 and 3, but, and this is a crucial point, in these immunization problems, we assume parallel shifts in the yield curve, that is all rates change by the same amount, and since we initially had a flat yield curve (since the yield to maturity of both bonds is 0.10) we assume the yield curve remains flat implying $R_{23}=R_{12}$.
Now, we want to minimize the change in $V$ with respect to changes in $R_{12}$. Take the derivate and set to 0 to minimize, to get
$$\frac{d V}{d R}= -x +\frac{y}{(1+R_{12})}=0.$$
This reduces to $$x=\frac{y}{(1+R_{12})^2}$$
Now, we don't know $R_{12}$, but we are thinking about small changes around the current interest rate, which is $10\%$, to that becomes $x=\frac{y}{(1.10)^2}$.
But we also know the value in period 2 has to be 1000, so using the formula for $V$ and substituting in the answer above for $x$ we know that
$$\frac{y}{(1.10)^2} (1.10) + \frac{y}{1.10} =10$$ where I've divided out the 100 face value. That simplifies to $2y=11$ and the answer follows.