A photon of light with wavelength 700 nm has energy of photon of light of what wavelength would correspond to energy 2E

@ sakshi 
the question is asking about photons , so, we can directly apply the following formula .. 
E= Hv here h is the planks constant , and v is the frequency . 
and as u knows v = c/ lambda 
where c is the speed of light and lambda is the wavelength . 
so, by plugging in the value of c in equation u will get 
E= HC/LAMBDA . 

now put the value of lambda and calculate E , 
E= 6.62*10^-34 *3*10^8 / 700*10^-9 
CALCULATE THE ABOVE AND U WILL GET E FROM THEIR . 
NOW, COMES TO THE 2ND PART OF QUESTION , 
ITS ASKING ABOUT CALCULATION OF LAMBDA WHEN 2E IS GIVEN 
SO, FORMULATE THE EQUATION 
2E = 2 HC/ LAMBDA 
AND FROM THE FIRST EQUATION U HAVE GOT THE VALUE OF E , 
MULTIPLY IT BY 2 , AND APPLY THAT HERE , 
U KNOWS THE REST THING 
h= 6.62*10^-34
c= 3*10^8 
and lambda u have to calculate .. 
put the below formula 
LAMBDA = HC/E 
AND CALCULATE IT ,