Answer
See the detailed answer below.
Work Step by Step
Let's assume that the actual wavelength of the proton is $\lambda_0$
We know that
$$E^2=p^2c^2+m^2c^4$$
So that
$$ p^2c^2=E^2-m^2c^4 $$
We also know that $E=KE+mc^2$, so
$$ p^2c^2=( KE+mc^2)^2-m^2c^4 $$
$$ p^2c^2= KE^2+2mc^2KE+m^2c^4 -m^2c^4 $$
$$ p^2c^2= KE^2+2mc^2KE \tag 1 $$
We know that the wavelength is given by
$$\lambda_0=\dfrac{h}{p}$$
Solving (1) for $p$ and then plugging it;
$$\lambda_0=\dfrac{h}{\sqrt{\dfrac{KE^2+2mc^2KE}{c^2}}}=\dfrac{hc}{KE\sqrt{1+\dfrac{2mc^2}{KE}}}$$
$$\lambda_0 = \dfrac{hc}{KE\sqrt{1+\dfrac{2mc^2}{KE}}}\tag 2$$
where $\lambda=\dfrac{hc}{KE}\tag 3$, where $\lambda$ is the approximate wavelength.
So that; $\lambda=1.01\lambda$ since $\lambda \gt \lambda_0$
Thus;
$$\lambda = 1.01\lambda_0$$
Plugging from (2);
$$\lambda=\dfrac{1.01hc}{KE\sqrt{1+\dfrac{2mc^2}{KE}}}\tag 2$$
Plugging from (3);
$$\dfrac{hc}{KE}=\dfrac{1.01hc}{KE\sqrt{1+\dfrac{2mc^2}{KE}}} $$
$$1=\dfrac{1.01 }{ \sqrt{1+\dfrac{2mc^2}{KE}}} $$
$$\sqrt{1+\dfrac{2mc^2}{KE}}= 1.01 $$
Squaring both sides;
$$ 1+\dfrac{2mc^2}{KE} = 1.01^2 $$
Solving for $KE$;
$$ {2mc^2} = (1.01^2-1)KE $$
$$KE=\dfrac{2mc^2}{1.01^2-1}$$
Plugging the known;
$$KE=\dfrac{2 \times 938.3\times 10^6}{1.01^2-1}=\color{red}{\bf 9.34\times 10^{10}}\;\rm eV$$
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Thus, the corresponding wavelength is given by
$$\lambda=\dfrac{hc}{KE}$$
Plugging the known;
$$\lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{9.34\times 10^{10}\times 1.6\times 10^{-19}}$$
$$\lambda=\color{red}{\bf 1.33\times 10^{-17}}\;\rm m$$
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and its speed is given by;
$$KE=mc^2\left[\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1\right]$$
Solving for $v$;
$$KE= \dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-mc^2 $$
$$KE+mc^2= \dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}} $$
$$\dfrac{KE+mc^2}{mc^2}= \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}} $$
$$\dfrac{mc^2}{KE+mc^2}= \sqrt{1-\dfrac{v^2}{c^2}} $$
$$\left(\dfrac{mc^2}{KE+mc^2}\right)^2= 1-\dfrac{v^2}{c^2 } $$
$$\dfrac{v^2}{c^2 } = 1- \left(\dfrac{mc^2}{KE+mc^2}\right)^2 $$
$$ v^2 = c^2\left[1 - \left(\dfrac{mc^2}{KE+mc^2}\right)^2 \right]$$
$$ v =\sqrt{ c^2\left[1 - \left(\dfrac{mc^2}{KE+mc^2}\right)^2 \right]}$$
$$ v =c\sqrt{ \left[1 - \left(\dfrac{mc^2}{KE+mc^2}\right)^2 \right]}$$
Plugging the known;
$$ v =c\sqrt{ \left[1 - \left(\dfrac{(938.3\times 10^6) }{(9.34\times 10^{10})+[(938.3\times 10^6) ]}\right)^2 \right]}$$
$$v=0.999951c=0.999951\times 3\times 10^8$$
$$v=\color{red}{\bf 2.9998\times 10^8}\;\rm m/s$$
